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ganeshie8
 one year ago
show that
\[\large \dfrac{(a,b,c)^2}{(a,b)(b,c)(c,a)}=\dfrac{[a,b,c]^2}{[a,b][b,c][c,a]}\]
where \((a,b,c)=\gcd(a,b,c)\)
\([a,b,c]=\text{lcm}(a,b,c)\)
\(a,b,c\in \mathbb{Z^{+}}\)
ganeshie8
 one year ago
show that \[\large \dfrac{(a,b,c)^2}{(a,b)(b,c)(c,a)}=\dfrac{[a,b,c]^2}{[a,b][b,c][c,a]}\] where \((a,b,c)=\gcd(a,b,c)\) \([a,b,c]=\text{lcm}(a,b,c)\) \(a,b,c\in \mathbb{Z^{+}}\)

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imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0m tryin hard but i haven't learnt such things..is this question related to number theory?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1ive got this instead :\ \(\Large \large \dfrac{(a,b,c)^2}{(a,b)(b,c)(c,a)}=\dfrac{[a,b][b,c][c,a]}{[a,b,c]^2 } \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let the prime factorizations of a, b, c be as follows: \(p_1^{a_1}...p_k^{a_k}\) \(p_1^{b_1}...p_k^{b_k}\) \(p_1^{c_1}...p_k^{c_k}\) Note that when we write it in this form some of the exponents might be zero since not all the primes have to be factors. But also note that at least one of \(a_i , b_i , c_i \) is nonzero because otherwise we wouldn't consider that prime at all. Now, note that \((a,b) = p_1^{min(a_1,b_1)}...p_k^{min(a_k,b_k)}\) and \((a,b,c) = p_1^{min(a_1,b_1,c_1)}...p_k^{min(a_k,b_k.c_k)}\) Similarly \( [a,b] = p_1^{max(a_1,b_1)}...p_k^{max(a_k,b_k)}\) and \([a,b,c] = p_1^{max(a_1,b_1,c_1)}...p_k^{max(a_k,b_k.c_k)}\) Using these, the equality can be reduced to \( 2min(a_i,b_i,c_i)  min(a_i,b_i)  min(b_i,c_i)  min(c_i,a_i)\) \( = 2max(a_i,b_i,c_i)  max(a_i,b_i)  max(b_i,c_i)  max(c_i,a_i)\) for each i That in turn can be reduced to \(max(a_i,b_i)  min(a_i,b_i) + max(b_i,c_i)  min(b_i,c_i) + max(c_i,a_i)  min(c_i,a_i) \) \(= 2[max(a_i,b_i,c_i)  min(a_i,b_i,c_i)]\) that is, \( a_i  b_i + b_ic_i + c_ia_i = 2[max(a_i,b_i,c_i)  min(a_i,b_i,c_i)]\) which is easily verified For example, WLOG assume \(c_i\) is the middle one, then, \(a_i  b_i + b_ic_i + c_ia_i = 2a_ib_i\) [\(c_i\) cancels out since in the sum it occurs twice with opposite signs] \(= 2[max(a_i,b_i)  min(a_i,b_i)]\) \(= 2[max(a_i,b_i,c_i)  min(a_i,b_i,c_i)]\) Thus for each i, we have \[ \frac{p_i^{2min(a_i,b_i,c_i)}}{p_i^{min(a_i,b_i)}p_i^{min(b_i,c_i)}p_i^{min(c_i,a_i)}} = \frac{p_i^{2max(a_i,b_i,c_i)}}{p_i^{max(a_i,b_i)}p_i^{max(b_i,c_i)}p_i^{max(c_i,a_i)}}\] Now multiplying both sides over \(i\) we get the desired equality.

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1i have reach this and went into a cycle :\ dw:1439110498143:dw

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439110656684:dw

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0I do not see how the second part be 1. Pairwise GCD of a,b,c will not be 1 unless they are coprime. 1/LCM is certainly smaller than 1 as LCM>=1 unless a,b,c=1.

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1that was pointless i'll try something else :)
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